VITEEE 2016 Paper

Let y denote the number of bacteria at any instant t • then according to the question

dy

dx

αy ⇒

dy

y

= kdt ... (i)
k is the constant of proportionality, taken to be + ve on integrating (i), we get
log y = kt + c ...(ii)
c is a parameter, let y0 be the initial number of bacteria
i.e., at t = 0 using this in (ii), c = log y0
⇒ log y = kt + log y0
⇒ log

y

y0

= kt ... (iii)
y = (y0+

10

100

y) =

11y0

10

, when t = 2
So, from (iii), we get log

11y0 10

y0

= k (2)
⇒ k =

1

2

log

11

10

... (iv)
Using (iv) in (iii) log

y

y0

=

1

2

(log

11

10

) t ... (v)
let the number of bacteria become 1, 00, 000 to 2,00,000 in t1 hours, i.e., y = 2y0 when t = t1 hours, from (v)
log

2y0

y0

=

1

2

(log

11

10

)t1 ⇒ t1 =

2log2

log 11 10

Hence, the reqd. no. of hours =

2log2

log( 11 10 )